Check sibling questions

Find derivative by first principle f(x) = xsinx [with Video] - Teachoo

Example 20 (ii) - Chapter 13 Class 11 Limits and Derivatives - Part 2

Example 20 (ii) - Chapter 13 Class 11 Limits and Derivatives - Part 3

This video is only available for Teachoo black users

This video is only available for Teachoo black users

Get live Maths 1-on-1 Classs - Class 6 to 12


Transcript

Example 20 Find the derivative of f(x) from the first principle, where f(x) is (ii) x sin x Given f (x) = x sin x We need to find Derivative of f(x) We know that f’(x) = lim┬(hβ†’0) 𝑓⁑〖(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯)γ€—/β„Ž Here, f (x) = x sin x So, f (x + h) = (x + h) sin (x + h) Putting values f’(x) =lim┬(hβ†’0) ((π‘₯ + β„Ž) sin⁑〖 (π‘₯ + β„Ž) βˆ’ π‘₯ sin⁑〖π‘₯ γ€— γ€—)/β„Ž Using sin (A + B) = sin A cos B + cos A sin B = lim┬(hβ†’0)⁑〖((π‘₯ + β„Ž)(sin⁑〖π‘₯ cosβ‘γ€–β„Ž +γ€– cos π‘₯〗⁑sinβ‘γ€–β„Ž γ€— )γ€— βˆ’ π‘₯ sin⁑π‘₯ γ€—)/β„Žγ€— = lim┬(hβ†’0)⁑〖(π‘₯(sin⁑〖π‘₯ cosβ‘γ€–β„Ž +γ€– cos〗⁑〖π‘₯ sinβ‘γ€–β„Ž) + β„Ž (sin⁑〖π‘₯ cosβ‘γ€–β„Ž +γ€– cos〗⁑〖π‘₯ sinβ‘γ€–β„Ž) βˆ’ π‘₯ sin⁑π‘₯ γ€— γ€— γ€— γ€— γ€— γ€— γ€— γ€—)/β„Žγ€— = lim┬(hβ†’0)⁑〖(π‘₯ sin⁑〖π‘₯ cosβ‘γ€–β„Ž + π‘₯ cos⁑〖π‘₯ sinβ‘γ€–β„Ž + β„Ž γ€–(sin〗⁑〖π‘₯ cosβ‘γ€–β„Ž + cos⁑〖π‘₯ sinβ‘γ€–β„Ž) βˆ’ π‘₯ sin⁑π‘₯ γ€— γ€— γ€— γ€— γ€— γ€— γ€— γ€—)/β„Žγ€— = lim┬(hβ†’0)⁑〖(π‘₯𝑠𝑖𝑛 π‘₯ cosβ‘γ€–β„Ž βˆ’ π‘₯ sin⁑〖π‘₯ + π‘₯ cos⁑〖π‘₯ sinβ‘γ€–β„Ž + β„Ž(sin⁑〖π‘₯ cosβ‘γ€–β„Ž + cos⁑〖π‘₯ sinβ‘β„Ž γ€— γ€—)γ€— γ€— γ€— γ€— γ€—)/β„Žγ€— = lim┬(hβ†’0)⁑〖(π‘₯𝑠𝑖𝑛 π‘₯ γ€–(cosγ€—β‘γ€–β„Ž βˆ’ 1)+ π‘₯ cos⁑〖π‘₯ sinβ‘γ€–β„Ž + β„Ž(sin⁑〖π‘₯ cosβ‘γ€–β„Ž + cos⁑〖π‘₯ sinβ‘β„Ž γ€— γ€—)γ€— γ€— γ€— γ€—)/β„Žγ€— = lim┬(hβ†’0)⁑((π‘₯ sin⁑〖π‘₯ (cosβ‘γ€–β„Ž βˆ’1)γ€— γ€—)/β„Ž+(π‘₯ cos⁑〖π‘₯ sinβ‘β„Ž γ€—)/β„Ž+(β„Ž (sin⁑〖π‘₯ cosβ‘γ€–β„Ž +cos⁑〖π‘₯ sinβ‘γ€–β„Ž)γ€— γ€— γ€— γ€—)/β„Ž) = lim┬(hβ†’0)⁑〖〖x sin〗⁑〖π‘₯ (cosβ‘γ€–β„Ž βˆ’1)γ€— γ€—/β„Ž+lim┬(hβ†’0) (π‘₯ cos⁑〖π‘₯ sinβ‘β„Ž γ€— )/h+lim┬(hβ†’0) (sin⁑〖π‘₯ cosβ‘γ€–β„Ž +cos⁑〖π‘₯ sinβ‘γ€–β„Ž )γ€— γ€— γ€— γ€— γ€— = – x sin x (π₯𝐒𝐦)┬(π‘β†’πŸŽ) ((πŸβˆ’γ€– 𝒄𝒐𝒔〗⁑〖𝒉)γ€—)/𝒉+π‘₯ cos⁑〖π‘₯ (π₯𝐒𝐦)┬(π‘β†’πŸŽ) π’”π’Šπ’β‘π’‰/𝐑+γ€— lim┬(hβ†’0) (sin⁑〖π‘₯ cosβ‘γ€–β„Ž +cos⁑〖π‘₯ sinβ‘γ€–β„Ž )γ€— γ€— γ€— γ€— = – x sin x (0) + x cos x (1) + ( sin x cos 0 + cos x sin 0) = 0 + x cos x + sin Γ— 1 + cos x Γ— 0 = 0 + x cos x + sin x + 0 = x cos x + sin x Hence f’ (x) = x cos x + sin x

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.